270385
The time period of an earth's satellite in circular orbit is independent of
1 the mass of the satellite
2 radius of its orbit
3 both the mass and radius of the orbit
4 neither the mass of the satellite nor the radius of its orbit
Explanation:
Gravitation
270491
The distance of Neptune and Saturn from the Sun are respectively \(10^{13}\) and \(10^{12}\) meters and their periodic times are respectively \(T_{n}\) and \(T_{s}\). If their orbits are circular, then the value of \(T_{n} / T_{s}\) is
270492
The Earth moves around the Sun in an elliptical orbit as shown in the figure. The ratio \(\frac{O A}{O B}=x\). Then, ratio of the speed of the Earth at \(B\) and at \(A\) is nearly
1 56
2 28
3 \(14 \sqrt{2}\)
4 7
Explanation:
From conservation of angular momentum
\(\mathrm{mv} \mathrm{r}=\text { Constant, } \mathrm{v}_{1} \mathrm{r}_{1}=\mathrm{v}_{2} \mathrm{r}_{2}\)
Q(119. The period of moon's rotation around the earth is nearly 29 days. If moon's mass were 2 fold its present value and all other things remain unchanged, the period of moon's rotation would be nearly (in days)
(a.) \(29 \sqrt{2}\)
(b.) \(29 / \sqrt{2}\)
(c.) \(29 \sqrt{3}\)
(d.) 29
Ans: d
Exp: Time period does not depend upon the mass of the satellite
Q(120. If the mass of earth were 2 times the present mass, the mass of the moon were half the present mass and the moon were revolving round the earth at the same present distance, the time period of revolution of the moon would be (in days)
(a.) 56
(b.) 28
(c.) \(14 \sqrt{2}\)
(d.) 7
Ans: c
Exp: From Kepler's 3rd law, \(T=2 \pi \sqrt{\frac{r^{3}}{G M}}\)
Gravitation
270533
In planetary motion, the areal velocity of position vector of a planet depends on angular velocity \((\omega)\) and the distance of the planet from sun (r). If so, the correct relation for areal velocity is
(2003E)
1 \(\frac{d A}{d t} \propto \omega r\)
2 \(\frac{d A}{d t} \propto \omega^{2} r\)
3 \(\frac{d A}{d t} \propto \omega r^{2}\)
4 \(\frac{d A}{d t} \propto \sqrt{\omega r}\)
Explanation:
\(\frac{d A}{d t}=\frac{L}{2 m} \Rightarrow \frac{d A}{d t} \propto v r \propto \omega r^{2}\)
270385
The time period of an earth's satellite in circular orbit is independent of
1 the mass of the satellite
2 radius of its orbit
3 both the mass and radius of the orbit
4 neither the mass of the satellite nor the radius of its orbit
Explanation:
Gravitation
270491
The distance of Neptune and Saturn from the Sun are respectively \(10^{13}\) and \(10^{12}\) meters and their periodic times are respectively \(T_{n}\) and \(T_{s}\). If their orbits are circular, then the value of \(T_{n} / T_{s}\) is
270492
The Earth moves around the Sun in an elliptical orbit as shown in the figure. The ratio \(\frac{O A}{O B}=x\). Then, ratio of the speed of the Earth at \(B\) and at \(A\) is nearly
1 56
2 28
3 \(14 \sqrt{2}\)
4 7
Explanation:
From conservation of angular momentum
\(\mathrm{mv} \mathrm{r}=\text { Constant, } \mathrm{v}_{1} \mathrm{r}_{1}=\mathrm{v}_{2} \mathrm{r}_{2}\)
Q(119. The period of moon's rotation around the earth is nearly 29 days. If moon's mass were 2 fold its present value and all other things remain unchanged, the period of moon's rotation would be nearly (in days)
(a.) \(29 \sqrt{2}\)
(b.) \(29 / \sqrt{2}\)
(c.) \(29 \sqrt{3}\)
(d.) 29
Ans: d
Exp: Time period does not depend upon the mass of the satellite
Q(120. If the mass of earth were 2 times the present mass, the mass of the moon were half the present mass and the moon were revolving round the earth at the same present distance, the time period of revolution of the moon would be (in days)
(a.) 56
(b.) 28
(c.) \(14 \sqrt{2}\)
(d.) 7
Ans: c
Exp: From Kepler's 3rd law, \(T=2 \pi \sqrt{\frac{r^{3}}{G M}}\)
Gravitation
270533
In planetary motion, the areal velocity of position vector of a planet depends on angular velocity \((\omega)\) and the distance of the planet from sun (r). If so, the correct relation for areal velocity is
(2003E)
1 \(\frac{d A}{d t} \propto \omega r\)
2 \(\frac{d A}{d t} \propto \omega^{2} r\)
3 \(\frac{d A}{d t} \propto \omega r^{2}\)
4 \(\frac{d A}{d t} \propto \sqrt{\omega r}\)
Explanation:
\(\frac{d A}{d t}=\frac{L}{2 m} \Rightarrow \frac{d A}{d t} \propto v r \propto \omega r^{2}\)
270385
The time period of an earth's satellite in circular orbit is independent of
1 the mass of the satellite
2 radius of its orbit
3 both the mass and radius of the orbit
4 neither the mass of the satellite nor the radius of its orbit
Explanation:
Gravitation
270491
The distance of Neptune and Saturn from the Sun are respectively \(10^{13}\) and \(10^{12}\) meters and their periodic times are respectively \(T_{n}\) and \(T_{s}\). If their orbits are circular, then the value of \(T_{n} / T_{s}\) is
270492
The Earth moves around the Sun in an elliptical orbit as shown in the figure. The ratio \(\frac{O A}{O B}=x\). Then, ratio of the speed of the Earth at \(B\) and at \(A\) is nearly
1 56
2 28
3 \(14 \sqrt{2}\)
4 7
Explanation:
From conservation of angular momentum
\(\mathrm{mv} \mathrm{r}=\text { Constant, } \mathrm{v}_{1} \mathrm{r}_{1}=\mathrm{v}_{2} \mathrm{r}_{2}\)
Q(119. The period of moon's rotation around the earth is nearly 29 days. If moon's mass were 2 fold its present value and all other things remain unchanged, the period of moon's rotation would be nearly (in days)
(a.) \(29 \sqrt{2}\)
(b.) \(29 / \sqrt{2}\)
(c.) \(29 \sqrt{3}\)
(d.) 29
Ans: d
Exp: Time period does not depend upon the mass of the satellite
Q(120. If the mass of earth were 2 times the present mass, the mass of the moon were half the present mass and the moon were revolving round the earth at the same present distance, the time period of revolution of the moon would be (in days)
(a.) 56
(b.) 28
(c.) \(14 \sqrt{2}\)
(d.) 7
Ans: c
Exp: From Kepler's 3rd law, \(T=2 \pi \sqrt{\frac{r^{3}}{G M}}\)
Gravitation
270533
In planetary motion, the areal velocity of position vector of a planet depends on angular velocity \((\omega)\) and the distance of the planet from sun (r). If so, the correct relation for areal velocity is
(2003E)
1 \(\frac{d A}{d t} \propto \omega r\)
2 \(\frac{d A}{d t} \propto \omega^{2} r\)
3 \(\frac{d A}{d t} \propto \omega r^{2}\)
4 \(\frac{d A}{d t} \propto \sqrt{\omega r}\)
Explanation:
\(\frac{d A}{d t}=\frac{L}{2 m} \Rightarrow \frac{d A}{d t} \propto v r \propto \omega r^{2}\)
270385
The time period of an earth's satellite in circular orbit is independent of
1 the mass of the satellite
2 radius of its orbit
3 both the mass and radius of the orbit
4 neither the mass of the satellite nor the radius of its orbit
Explanation:
Gravitation
270491
The distance of Neptune and Saturn from the Sun are respectively \(10^{13}\) and \(10^{12}\) meters and their periodic times are respectively \(T_{n}\) and \(T_{s}\). If their orbits are circular, then the value of \(T_{n} / T_{s}\) is
270492
The Earth moves around the Sun in an elliptical orbit as shown in the figure. The ratio \(\frac{O A}{O B}=x\). Then, ratio of the speed of the Earth at \(B\) and at \(A\) is nearly
1 56
2 28
3 \(14 \sqrt{2}\)
4 7
Explanation:
From conservation of angular momentum
\(\mathrm{mv} \mathrm{r}=\text { Constant, } \mathrm{v}_{1} \mathrm{r}_{1}=\mathrm{v}_{2} \mathrm{r}_{2}\)
Q(119. The period of moon's rotation around the earth is nearly 29 days. If moon's mass were 2 fold its present value and all other things remain unchanged, the period of moon's rotation would be nearly (in days)
(a.) \(29 \sqrt{2}\)
(b.) \(29 / \sqrt{2}\)
(c.) \(29 \sqrt{3}\)
(d.) 29
Ans: d
Exp: Time period does not depend upon the mass of the satellite
Q(120. If the mass of earth were 2 times the present mass, the mass of the moon were half the present mass and the moon were revolving round the earth at the same present distance, the time period of revolution of the moon would be (in days)
(a.) 56
(b.) 28
(c.) \(14 \sqrt{2}\)
(d.) 7
Ans: c
Exp: From Kepler's 3rd law, \(T=2 \pi \sqrt{\frac{r^{3}}{G M}}\)
Gravitation
270533
In planetary motion, the areal velocity of position vector of a planet depends on angular velocity \((\omega)\) and the distance of the planet from sun (r). If so, the correct relation for areal velocity is
(2003E)
1 \(\frac{d A}{d t} \propto \omega r\)
2 \(\frac{d A}{d t} \propto \omega^{2} r\)
3 \(\frac{d A}{d t} \propto \omega r^{2}\)
4 \(\frac{d A}{d t} \propto \sqrt{\omega r}\)
Explanation:
\(\frac{d A}{d t}=\frac{L}{2 m} \Rightarrow \frac{d A}{d t} \propto v r \propto \omega r^{2}\)
